How do you use the convergence tests, determine whether the given series converges #sum (7-sin(n^2))/n^2+1# from n to infinity?

Question

William Abdalla · Accepted Answer

If the general term is with the #1#, the series is divergent, if the #1# is not in the term, then follow me: The function sinus has a range #[-1,1]#, so it doesn't influence the character of the series. So: #(7-sin(n^2))/n^2+1~1/n^2# (#~# means asyntotic). Since the function #1/n^2# is convergent, so it is the given one.

Luke Alvarez · Answer

First do the limit test for convergence, and check the limit, as the limit needs to equal 0 for a series to be considered convergent, if the limit is not 0, it is then divergent.

Thus:

#lim_(n -> oo) (7 - sin(n^2))/n^2 + 1 = 0 + 1 = 1#

Therefore we can conclude that the series does not converge, and therefore is Divergent.

Isabella Ackerman · Answer

undefined To determine the convergence of the series $\sum_{n=1}^\infty \frac{7 - \sin(n^2)}{n^2 + 1}$, we can use the Limit Comparison Test.

Let's denote the series as $a_n = \frac{7 - \sin(n^2)}{n^2 + 1}$.

Consider a simpler series $b_n = \frac{1}{n^2}$, which is a well-known convergent p-series with $p = 2$.

Now, we find the limit of the ratio $L$ as $n$ approaches infinity:

$$L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{7 - \sin(n^2)}{n^2 + 1}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{7 - \sin(n^2)}{n^2 + 1} \cdot n^2 = \lim_{n \to \infty} \left(7n^2 - \frac{\sin(n^2)}{n^2 + 1} \cdot n^2\right)$$

Since $\lim_{n \to \infty} \frac{\sin(n^2)}{n^2} = 0$, we can simplify $L$ to:

$$L = \lim_{n \to \infty} \left(7n^2 - \frac{\sin(n^2)}{n^2 + 1} \cdot n^2\right) = \lim_{n \to \infty} 7n^2 = \infty$$

Since $L = \infty$, the Limit Comparison Test implies that either both series converge or both diverge. Since $b_n$ is known to converge, the original series $\sum_{n=1}^\infty \frac{7 - \sin(n^2)}{n^2 + 1}$ also converges.